6.2.1 Equations of planes

Introduction to Equations of Planes

A plane in 3D space can be represented in different forms. Two common representations are:

Vector Form:

\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}

where:

$\mathbf{r}$ is the position vector of any point on the plane.
$\mathbf{a}$ is the position vector of a fixed point on the plane.
$\mathbf{b}$ and $\mathbf{c}$ are non-parallel direction vectors lying on the plane.
$\lambda$ and $\mu$ are scalar parameters.

Cartesian Form:

ax + by + cz = d

where:

$(a, b, c)$ is the normal vector perpendicular to the plane.
dd is a constant.

Deriving the Cartesian Form from the Vector Form

Step 1: Write the Vector Form

\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}

Step 2: Identify Two Direction Vectors

Let:

\mathbf{r} - \mathbf{a} = \lambda \mathbf{b} + \mu \mathbf{c}

The vectors $\mathbf{b}$ and $\mathbf{c}$ lie on the plane.

Step 3: Use the Normal Vector

The normal vector $\mathbf{n}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$ , so:

\mathbf{n} = \mathbf{b} \times \mathbf{c}

Step 4: Find the Cartesian Form

If $\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $\mathbf{a} = \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}$ , then:

a(x - x_1) + b(y - y_1) + c(z - z_1) = 0

which simplifies to:

ax + by + cz = d

where $d = ax_1 + by_1 + cz_1$

Worked Examples

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Example 1: Find the Vector Equation of a Plane

Find the vector equation of the plane passing through $(1, 2, 3)$ and containing the direction vectors $\mathbf{b} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$ and $c=(112)\mathbf{c} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

Step 1**: Write the vector equation:**

\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}

where $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$

Step 2**: Substitute values:**

\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}

Result:

\mathbf{r} = \begin{pmatrix} 1 + 4\lambda + \mu \\ 2 - 2\lambda + \mu \\ 3 + \lambda + 2\mu \end{pmatrix}

infoNote

Example 2: Convert Vector Form to Cartesian Form

Convert the plane:

\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}

into Cartesian form.

Step 1: Find the normal vector: Compute the cross-product of $\mathbf{b} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$ and $\mathbf{c} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -2 & 1 \\ 1 & 1 & 2 \end{vmatrix}

= \mathbf{i}((-2)(2) - (1)(1)) - \mathbf{j}((4)(2) - (1)(1)) + \mathbf{k}((4)(1) - (-2)(1))

\mathbf{n} = \begin{pmatrix} -5 \\ -7 \\ 6 \end{pmatrix}

Step 2: Substitute into the plane equation:

Let $(x_1, y_1, z_1) = (1, 2, 3)$

-5(x - 1) - 7(y - 2) + 6(z - 3) = 0

Step 3**: Simplify:**

-5x - 7y + 6z = -5(-1) - 7(-2) + 6(-3) = 11

Result:

-5x - 7y + 6z = 11

infoNote

Example 3: Plane Equation from Points

Find the Cartesian equation of the plane passing through $(1, 2, 3)$ , $(2, 0, 1)$ , and $(0, 1, 4)$

Step 1: Find two direction vectors:

\mathbf{b} = \begin{pmatrix} 2 - 1 \\ 0 - 2 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix}

\mathbf{c} = \begin{pmatrix} 0 - 1 \\ 1 - 2 \\ 4 - 3 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}

Step 2: Find the normal vector:

\mathbf{n} = \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -2 \\ -1 & -1 & 1 \end{vmatrix}

\mathbf{n} = \mathbf{i}((-2)(1) - (-1)(-2)) - \mathbf{j}((1)(1) - (-1)(-2)) + \mathbf{k}((1)(-1) - (-1)(-2))

\mathbf{n} = \begin{pmatrix} -4 \\ -3 \\ -3 \end{pmatrix}

Step 3: Write the Cartesian equation:

Use $(1, 2, 3)$

-4(x - 1) - 3(y - 2) - 3(z - 3) = 0

Step 4: Simplify:

-4x - 3y - 3z = -4(-1) - 3(-2) - 3(-3) = -22

Result:

−4x - 3y - 3z = -22

Note Summary

infoNote

Common Mistakes:

Incorrect cross-product calculations: Ensure the determinant is calculated carefully to find the normal vector.
Mixing up direction and normal vectors: Direction vectors lie in the plane, while the normal vector is perpendicular to the plane.
Forgetting to substitute a point for dd: In Cartesian form, $d = ax_1 + by_1 + cz_1$ for a point $(x_1, y_1, z_1)$
Confusing the vector and Cartesian forms: The vector form involves direction vectors, while the Cartesian form uses the normal vector.
Omitting terms in simplifications: Be careful to expand and simplify equations correctly.

infoNote

Key Formulas:

Vector Form of a Plane:

\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}

Cartesian Form of a Plane:

ax + by + cz = d

Normal Vector Calculation:

\mathbf{n} = \mathbf{b} \times \mathbf{c}

Point Substitution for dd:

d = ax_1 + by_1 + cz_1

Equations of planes Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

6.2.1 Equations of planes

Introduction to Equations of Planes

Vector Form:

Cartesian Form:

Deriving the Cartesian Form from the Vector Form

Step 1: Write the Vector Form

Step 2: Identify Two Direction Vectors

Step 3: Use the Normal Vector

Step 4: Find the Cartesian Form

Worked Examples

Example 1: Find the Vector Equation of a Plane

Example 2: Convert Vector Form to Cartesian Form

Example 3: Plane Equation from Points

Note Summary

Common Mistakes:

Key Formulas:

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